Let's solve the equation:

$x - 5 = \sqrt{7 - x}$

Step 1: Eliminate the square root

To eliminate the square root, square both sides of the equation:

$(x - 5)^2 = (\sqrt{7 - x})^2$

This simplifies to:

$(x - 5)^2 = 7 - x$

Step 2: Expand the left-hand side

Now, expand $(x - 5)^2$:

$(x - 5)(x - 5) = x^2 - 10x + 25$

So the equation becomes:

$x^2 - 10x + 25 = 7 - x$

Step 3: Rearrange the terms

Move all terms to one side:

$x^2 - 10x + 25 + x - 7 = 0$

Simplify:

$x^2 - 9x + 18 = 0$

Step 4: Solve the quadratic equation

We now solve the quadratic equation $x^2 - 9x + 18 = 0$. Use the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Here, $a = 1$, $b = -9$, and $c = 18$. Substitute these into the formula:

$x = \frac{-(-9) \pm \sqrt{(-9)^2 - 4(1)(18)}}{2(1)}$ $x = \frac{9 \pm \sqrt{81 - 72}}{2}$ $x = \frac{9 \pm \sqrt{9}}{2}$ $x = \frac{9 \pm 3}{2}$

Step 5: Find the two possible solutions

Now, calculate the two possible values for $x$:

1. $x = \frac{9 + 3}{2} = \frac{12}{2} = 6$
2. $x = \frac{9 - 3}{2} = \frac{6}{2} = 3$

Step 6: Check the solutions

We need to check both solutions in the original equation to see if they are valid.

For $x = 6$: $6 - 5 = \sqrt{7 - 6}$ $1 = \sqrt{1}$ This is true, so $x = 6$ is a valid solution.

For $x = 3$: $3 - 5 = \sqrt{7 - 3}$ $-2 = \sqrt{4}$ $-2 = 2$ This is false, so $x = 3$ is not a valid solution.

Final Answer:

The only valid solution is: $x = 6$

Would you like more details or have any other questions?

Here are 5 related questions for further exploration:

1. How do we handle extraneous solutions in equations with square roots?
2. What is the method for solving quadratic equations by factoring?
3. What is the quadratic formula, and when should it be used?
4. How do we simplify equations involving square roots before squaring both sides?
5. How can we check for extraneous solutions when solving rational equations?

Tip: Always check your solutions in the original equation, especially when dealing with square roots or other non-linear transformations, to avoid extraneous solutions.